Number of subarrays having sum less than K

Given an array of numbers and a number k, find the number of subarrays having sum less than k. We may assume that there is no overflow.

Examples :

Input : arr[] = {2, 5, 6}
        K = 10
Output : 4
The subarrays are {2}, {5}, {6} and
{2, 5},

Input : arr[] = {1, 11, 2, 3, 15}
        K = 10
Output : 4
{1}, {2}, {3} and {2, 3}

A simple solution is to generate all subarrays of the array and then count the number of arrays having sum less than K.
Below is the implementation of above approach :

C++


[sourcecode language=”CPP”]
// CPP program to count
// subarrays having sum
// less than k.
#include <bits/stdc++.h>
using namespace std;

// Function to find number
// of subarrays having sum
// less than k.
int countSubarray(int arr[],
int n, int k)
{
int count = 0;

for (int i = 0; i < n; i++)
{
int sum = 0;
for (int j = i; j < n; j++)
{

// If sum is less than k
// then update sum and
// increment count
if (sum + arr[j] < k)
{
sum = arr[j] + sum;
count++;
}
else
{
break;
}
}
}

return count;
}

// Driver Code
int main()
{
int array[] = {1, 11, 2, 3, 15};
int k = 10;
int size = sizeof(array) / sizeof(array[0]);
int count = countSubarray(array, size, k);
cout << count << "\n";
}

[/sourcecode]

Java


[sourcecode language=”Java” highlight=”8-36″]
// Java program to count subarrays
// having sum less than k.
import java.io.*;

class GFG
{

// Function to find number of
// subarrays having sum less than k.
static int countSubarray(int arr[],
int n, int k)
{
int count = 0;

for (int i = 0; i < n; i++)
{
int sum = 0;
for (int j = i; j < n; j++)
{

// If sum is less than
// k then update sum and
// increment count
if (sum + arr[j] < k)
{
sum = arr[j] + sum;
count++;
}
else
{
break;
}
}
}
return count;
}

// Driver Code
public static void main (String[] args)
{
int array[] = {1, 11, 2, 3, 15};
int k = 10;
int size = array.length;
int count = countSubarray(array, size, k);
System.out.println(count);
}
}

// This code is contributed by Sam007

[/sourcecode]

Python3


[sourcecode language=”Python3″]
# python program to count subarrays
# having sum less than k.

# Function to find number of subarrays
# having sum less than k.
def countSubarray(arr, n, k):
count = 0

for i in range(0, n):
sum = 0;
for j in range(i, n):

# If sum is less than k
# then update sum and
# increment count
if (sum + arr[j] < k):
sum = arr[j] + sum
count+=1
else:
break
return count;

# Driver Code
array = [1, 11, 2, 3, 15]
k = 10
size = len(array)
count = countSubarray(array, size, k);
print(count)

# This code is contributed by Sam007
[/sourcecode]

C#


[sourcecode language=”CSHARP” highlight=”8-37″]
// C# program to count subarrays
// having sum less than k.
using System;

class GFG
{

// Function to find number
// of subarrays having sum
// less than k.
static int countSubarray(int []arr,
int n, int k)
{
int count = 0;

for (int i = 0; i < n; i++)
{
int sum = 0;
for (int j = i; j < n; j++)
{

// If sum is less than k
// then update sum and
// increment count
if (sum + arr[j] < k)
{
sum = arr[j] + sum;
count++;
}
else
{
break;
}
}
}
return count;
}

// Driver Code
public static void Main (String[] args)
{
int []array = {1, 11, 2, 3, 15};
int k = 10;
int size = array.Length;
int count = countSubarray(array, size, k);
Console.WriteLine(count);
}
}

// This code is contributed by Sam007

[/sourcecode]

PHP


[sourcecode language=”php”]

<?php
// PHP program to count
// subarrays having sum
// less than k.

// Function to find number
// of subarrays having sum
// less than k.
function countSubarray($arr, $n, $k)
{
$count = 0;

for ($i = 0; $i < $n; $i++)
{
$sum = 0;
for ($j = $i; $j < $n; $j++)
{

// If sum is less than
// k then update sum and
// increment count
if ($sum + $arr[$j] < $k)
{
$sum = $arr[$j] + $sum;
$count++;
}
else
{
break;
}
}
}

return $count;
}

// Driver Code
$array = array(1, 11, 2, 3, 15);
$k = 10;
$size = sizeof($array);
$count = countSubarray($array, $size, $k);
echo $count , "\n";

// This code is contributed by ajit
?>

[/sourcecode]

Output :

4


Time complexity :
O(n^2).

An efficient solution is based on sliding window technique that can be used to solve the problem. We use two pointers start and end to represent starting and ending point of sliding window. (Not that we need to find contiguous parts).

Initially both start and end point to the beginning of the array, i.e. index 0. Now, let’s try to add a new element el. There are two possible conditions.

1st case :
If sum is less than k, increment end by one position. So contiguous arrays this step produce are (end – start). We also add el to previous sum. There are as many such arrays as the length of the window.

2nd case :
If sum becomes greater than or equal to k, this means we need to subtract starting element from sum so that the sum again becomes less than k. So we adjust the window’s left border by incrementing start.

We follow the same procedure until end < array size.

Implementation:

C++


[sourcecode language=”CPP”]
// CPP program to count
// subarrays having sum
// less than k.
#include <bits/stdc++.h>
using namespace std;

// Function to find number
// of subarrays having sum
// less than k.
int countSubarrays(int arr[],
int n, int k)
{
int start = 0, end = 0,
count = 0, sum = arr[0];

while (end < n)
{

// If sum is less than k,
// move end by one position.
// Update count and sum
// accordingly.
if (sum < k)
{
end++;
count += end – start;

// For last element,
// end may become n
if (end < n)
sum += arr[end];
}

// If sum is greater than or
// equal to k, subtract
// arr[start] from sum and
// decrease sliding window by
// moving start by one position
else
{
sum -= arr[start];
start++;
}
}

return count;
}

// Driver Code
int main()
{
int array[] = {1, 11, 2, 3, 15};
int k = 10;
int size = sizeof(array) / sizeof(array[0]);
cout << countSubarrays(array, size, k);
}

[/sourcecode]

Java


[sourcecode language=”Java” highlight=”9-49″]
// Java program to count
// subarrays having sum
// less than k.
import java.io.*;

class GFG
{

// Function to find number
// of subarrays having sum
// less than k.
static int countSubarray(int arr[],
int n, int k)
{
int start = 0, end = 0;
int count = 0, sum = arr[0];

while (end < n)
{

// If sum is less than k,
// move end by one position.
// Update count and sum
// accordingly.
if (sum < k)
{
end++;
count += end – start;

// For last element,
// end may become n.
if (end < n)
sum += arr[end];
}

// If sum is greater than or
// equal to k, subtract
// arr[start] from sum and
// decrease sliding window by
// moving start by one position
else
{
sum -= arr[start];
start++;
}
}

return count;
}

// Driver Code
public static void main (String[] args)
{
int array[] = {1, 11, 2, 3, 15};
int k = 10;
int size = array.length;
int count = countSubarray(array, size, k);
System.out.println(count);
}
}

// This code is contributed by Sam007

[/sourcecode]

C#


[sourcecode language=”CSHARP” highlight=”9-49″]
// C# program to count
// subarrays having sum
// less than k.
using System;

class GFG
{

// Function to find number
// of subarrays having sum
// less than k.
static int countSubarray(int []arr,
int n, int k)
{
int start = 0, end = 0;
int count = 0, sum = arr[0];

while (end < n)
{

// If sum is less than k,
// move end by one position.
// Update count and sum
// accordingly.
if (sum < k)
{
end++;
count += end – start;

// For last element,
// end may become n.
if (end < n)
sum += arr[end];
}

// If sum is greater than or
// equal to k, subtract
// arr[start] from sum and
// decrease sliding window by
// moving start by one position
else
{
sum -= arr[start];
start++;
}
}

return count;
}

// Driver Code
public static void Main (String[] args)
{
int []array = {1, 11, 2, 3, 15};
int k = 10;
int size = array.Length;
int count = countSubarray(array, size, k);
Console.WriteLine(count);
}
}

// This code is contributed by Sam007

[/sourcecode]

PHP


[sourcecode language=”php”]

<?php
// PHP program to count
// subarrays having sum
// less than k.

// Function to find number
// of subarrays having sum
// less than k.
function countSubarrays($arr,
$n, $k)
{
$start = 0;
$end = 0;
$count = 0;
$sum = $arr[0];

while ($end < $n)
{

// If sum is less than k,
// move end by one position.
// Update count and sum
// accordingly.
if ($sum < $k)
{
$end++;
$count += $end – $start;

// For last element,
// end may become n
if ($end < $n)
$sum += $arr[$end];
}

// If sum is greater than or
// equal to k, subtract
// arr[start] from sum and
// decrease sliding window by
// moving start by one position
else
{
$sum -= $arr[$start];
$start++;
}
}

return $count;
}

// Driver Code
$array =array (1, 11, 2, 3, 15);
$k = 10;
$size = sizeof($array) ;
echo countSubarrays($array, $size, $k);

// This code is contributed by ajit
?>

[/sourcecode]

Output :

4

Time complexity : O(n).



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Improved By : Sam007, jit_t, Ita_c



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